If Sn=∑r=1n tr=16n2n2+9n+13, then ∑r=1n trequals

We have tn=Sn−Sn−1∀n≥2∴ tn=16n2n2+9n+13−16(n−1)2(n−1)2 =162n3−(n−1)3+9n2−(n−1)2 =166n2−6n+2+9(2n−1)+13 =166n2+12n+6 =(n+1)2Also, t1=S1=4=(1+1)2∴ ∑r=1n tr=∑r=1n (r+1)=12(n+1)(n+2)−1=12n(n+3)

If Sn=∑r=1n tr=16n2n2+9n+13, then ∑r=1n trequals

We have tn=Sn−Sn−1∀n≥2∴ tn=16n2n2+9n+13−16(n−1)2(n−1)2 =162n3−(n−1)3+9n2−(n−1)2 =166n2−6n+2+9(2n−1)+13 =166n2+12n+6 =(n+1)2Also, t1=S1=4=(1+1)2∴ ∑r=1n tr=∑r=1n (r+1)=12(n+1)(n+2)−1=12n(n+3)

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If $S_{n} = \sum_{r = 1}^{n} t_{r} = \frac{1}{6} n (2 n^{2} + 9 n + 13)$ , then $\sum_{r = 1}^{n} \sqrt{t_{r}}$ equals

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$\frac{1}{2} n (n + 1)$

$\frac{1}{2} n (n + 2)$

$\frac{1}{2} n (n + 3)$

$\frac{1}{2} n (n + 5)$

answer is C.

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Detailed Solution

We have $t_{n} = S_{n} - S_{n - 1} \forall n \geq 2$
$\begin{array}{l} ∴ t_{n} = \frac{1}{6} n (2 n^{2} + 9 n + 13) - \frac{1}{6} (n - 1) \{(2 (n - 1)^{2} \\ = \frac{1}{6} [2 (n^{3} - (n - 1)^{3}) + 9 (n^{2} - (n - 1)^{2}) \\ = \frac{1}{6} [6 n^{2} - 6 n + 2 + 9 (2 n - 1) + 13] \\ = \frac{1}{6} (6 n^{2} + 12 n + 6) \\ = (n + 1)^{2} \end{array}$
Also, $t_{1} = S_{1} = 4 = (1 + 1)^{2}$
$∴ \sum_{r = 1}^{n} \sqrt{t_{r}} = \sum_{r = 1}^{n} (r + 1) = \frac{1}{2} (n + 1) (n + 2) - 1 = \frac{1}{2} n (n + 3)$